Question

A manufacturer uses two machines to drill holes in pieces of sheet metal used in engine construction. The workers who attach the sheet metal to the engine become inspectors in that they reject sheets so poorly drilled that they cannot be attached. The production manager is interested in knowing whether one machine is produces more defective drillings than the other machine. As an experiment, employee mark the sheets so that the manager can determine which machine was used to drill the holes. A random sample of 191 sheets of metal drilled by machine 1 is taken, and 38 of the sheets are defective. A random sample of 202 sheets of metal drilled by machine 2 is taken, and 21 of the sheets are defective.

Use alpha = .05 to determine whether

a. There is a significant difference in the proportion of sheets drilled with defective holes between machine 1 and machine 2.

b. Machine 1 produces 8% more defective holes than machine 2.   

Answer 1

Solution:

a)

For sample 1, we have that the sample size is N1= 191​, the number of favorable cases is X1 ​=38, so then the sample proportion is \hat p1 = {X1}/{N1} = { 38}/{ 191} =0.199

For sample 2, we have that the sample size is N2 ​=202, the number of favorable cases is X2 = 21, so then the sample proportion is \hat p2 = {X_2}//{N_2} = { 21}/{ 202}​=0.104

The value of the pooled proportion is computed as bar p = { X1+X2}/{N1+N2}= { 38 + 21}/{ 191+202} =0.1501

Also, the given significance level is alpha =0.05.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p1​=p2​

Ha:p1​̸​=p2​

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is zc ​=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

z = hat p1 - hat p2}/{sqrt{ bar p(1-bar p)(1/n1 + 1/n2)}} = { 0.199 - 0.104}/{sqrt{ 0.1501(1-0.1501)(1/191 + 1/202)}} =2.635

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=2.635>zc​=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p =0.0084, and since p=0.0084<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p1​ is different than p2​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval for p1​−p2​ is: 0.024 <p1​−p2​<0.166.

Graphically

b)

For sample 1, we have that the sample size is N1= 191​, the number of favorable cases is X1 ​=38, so then the sample proportion is \hat p1 = {X1}/{N1} = { 38}/{ 191} =0.199

For sample 2, we have that the sample size is N2 ​=202, the number of favorable cases is X2 = 21, so then the sample proportion is \hat p2 = {X_2}//{N_2} = { 21}/{ 202}​=0.104

The value of the pooled proportion is computed as bar p = { X1+X2}/{N1+N2}= { 38 + 21}/{ 191+202} =0.1501

Also, the given significance level is alpha =0.05.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Ho:p1​=p2​

Ha: Ha:p1​<p2​( machin 1 produse 8% more defective than meachin 2)

This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a left-tailed test is zc =−1.64.

The rejection region for this left-tailed test is R={z:z<−1.64}

(3) Test Statistics

The z-statistic is computed as follows:

z = hat p1 - hat p2}/{sqrt{ bar p(1-bar p)(1/n1 + 1/n2)}} = { 0.199 - 0.104}/{sqrt{ 0.1501(1-0.1501)(1/191 + 1/202)}} =2.635

(4) Decision about the null hypothesis

Since it is observed that z =2.635≥zc​=−1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p =0.9958, and since p=0.9958≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p1​ is less than p2​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval for p1​−p2​ is: 0.024<p1​−p2​<0.166.

meachin 1 do not produse more defective than machine 2

Graphically

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